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How do I evaluate something like 3070^2-3069^2?

This looks pretty daunting, but the key to this sort of a problem, which has come up a couple of times in previous PAT tests, is to recognise that this expression is actually the difference of two squares. It can be generalised to the problem of evaluating:

x2-y2

which has as its solution:

x2-y2 = (x+y)(x-y)

so in this case:

30702-30692 = (3070+3069)(3070-3069)=(6139)(1)=6139

Answered by
Sam C.

Studies MEng Engineering Science at Oxford, New College

What is the sum of the series 2/3 − 2/9 + 2/27 − ....? (PAT Q1 2013)

This is the sum of a geometric series with an infinite number of terms.

First, find the common ratio:

(-2/9)/(2/3) = -1/3 , (2/27)/(-2/9) = -1/3

Therefore, the common ratio is -1/3

-1

The general formula for a sum of an infinite geometric series is a/(1-r) where a is the first number in the sequence and r is the common ratio.

So, substituting in the numbers, the sum of the series = (2/3)/(1-[-1/3]) = 1/2

Answered by
Kirill M.

Studies Physics at Oxford, St Catherine's College

If a bullet is shot into a suspended mass and does not pass through why can you not just use conservation of energy?

Energy will be lost due to the plastic deformation of the mass as the bullet enters the mass. However momentum will always be conserved, so you can use this to find the speed of the mass when the bullet enters.

Answered by
Dominic D.

Studies Physics at Oxford, Keble College

How are you to find 2007^2 − 2006^2 without a calculator?

Like with many questions in the PAT test once you know what to do it is quite easy! To solve you can use a difference of two squares i.e. (2007-2006)(2007+2006). As the first bracket gives 1 and the second gives 4013 (by arithmetic). Therefore you get 4013*1 = 4013

Answered by
Dominic D.

Studies Physics at Oxford, Keble College

How do I approximate an irrational square root without using a calculator?

Although this might look like a smart trick at the beginning, the truth is that as a physicist this kind of approximations turn out to be useful very often. This is quite likely to come up in the PAT as well!

It is important to get used to understanding symbolic expressions (no numbers!), so I will first derive the maths and then apply the resulting expression to a couple of illustrative examples.

Our aim is to obtain an approximate value for the inexact square root x of an arbitrary number n, i.e. to solve x = n1/2. In other words, we want to solve the equation f(x) = x2 = n. Suppose we know the exact root x0 of another number n0 which is very close to n, i.e. f(x0) = x02 = n0 (this is key!). We can do a Taylor series expansion of f(x) about x0 up to first order (first derivative f'(x)), which is simply:

We are almost done! We know all the values of n, n0 and x0 and just need to solve for x. Rearranging:

x = x0 + (n - n0)/(2x0)

That's it! Remember that this expression is just an approximation for the actual square root of n, but a quite good one the closer n0 is to n.

Let's plug in some numbers. Suppose you were asked to approximate the square root of n = 66. Despite the solution being irrational, there is a close number n0 = 64 with a nice exact square root x0 = 8. Using the expression above we find:

x = 8 + (66 - 64)/(2*8) = 8 + 2/16 = 8 + 1/8 = 8 + 0.125 = 8.125

How good is this? It's just 0.01% off the actual value 8.12404...!

What if we chose n0 = 81 instead, the square root of which is x0 = 9? This would give x = 8.167 - worse, but still a reasonable result considering how far 81 is from 66. Never forget to quote both the positive and negative square roots in your final solution!

Would you be able to derive a similar expression for approximating a cubic root?

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2016-12-05T16:44:39Z

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